ECE | 232 | Circuits and Systems 2 | lecture notes | 2

013213215X ism02.pdf

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Page 14

2–14

CHAPTER 2. Circuit Elements

across the left branch is 5 V, so 20 +

= 5 V, and

15 V, where

is positive at the top. Note that the current through the 20 V source must be

9 A, ﬂowing from top to bottom, and the current through the

is 6 A ﬂowing

from top to bottom. Let’s ﬁnd the power associated with the left and middle

branches:

= (9)(

15) =

135 W

20V

= (9)(20) = 180 W

(6)(

5) = 27 W

= (6)(0

5) = 3 W

Since there is only one component left, we can ﬁnd the total power:

total

135 + 180 + 27 + 3 +

= 75 +

must equal

75 W.

Therefore,



dev



abs

= 210 W

P 2.9

[a]

Yes, each of the voltage sources can carry the current required by the

interconnection, and each of the current sources can carry the voltage

drop required by the interconnection. (Note that

8 A.)

[b]

No, because the voltage drop between the top terminal and the bottom

terminal cannot be determined. For example, deﬁne

, and

shown:

The voltage drop across the left branch, the center branch, and the right

branch must be the same, since these branches are connected at the same

two terminals. This requires that

20 +

+ 100 =

But this equation has three unknown voltages, so the individual voltages

cannot be determined, and thus the power of the sources cannot be

determined.

P 2.10

[a]

obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,

mechanical, photocopying, recording, or likewise. For information

regarding permission(s), write to: Rights and Permissions Department,

Pearson Education, Inc., Upper Saddle River, NJ 07458.

Page 15

Problems

2–15

[b]

no-load voltage of battery

internal resistance of battery

resistance of wire between battery and switch

resistance of wire between switch and lamp A

resistance of lamp A

resistance of lamp B

resistance of wire between lamp A and lamp B

resistance of frame between battery and lamp A

resistance of frame between lamp A and lamp B

switch

P 2.11

Since we know the device is a resistor, we can use Ohm’s law to calculate the

resistance. From Fig. P2.11(a),

Using the values in the table of Fig. P2.11(b),

108

004

002

108

004

162

006

= 27 kΩ

Note that this value is found in Appendix H.

P 2.12

The resistor value is the ratio of the power to the square of the current:

. Using the values for power and current in Fig. P2.12(b),

(50

)

(100

)

(150

)

(200

)

137

(250

)

198

(300

)

2 MΩ

Note that this is a value from Appendix H.

P 2.13

Since we know the device is a resistor, we can use the power equation. From

Fig. P2.13(a),

obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,

mechanical, photocopying, recording, or likewise. For information

regarding permission(s), write to: Rights and Permissions Department,

Pearson Education, Inc., Upper Saddle River, NJ 07458.

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